Graph f(x)=2x^5-3x^4-11x^3+9x^2+15x.

Compute f(f(-1)).

Factor f and find it's zeroes.

Substitiute x = -1

f(-1) = 0

f(f(-1)) = 0 because f(0) = 0

So now you know that x is a factor and (x + 1) is a factor

2x^5-3x^4-11x^3+9x^2+15x = x(2x^4 -3x^3 -11x^2 + 9x + 15)

= x[ 2x^4 + 2x^3 -5 x^3 - 5 x^2 - 6x^2 - 6x + 15 x +15)

= x[(2x^3(x + 1) - 5x^2(x + 1) -6x (x + 1) +15 (x +1)]

= x(x+1) ( 2x^3 - 5x^2 - 6x + 15)

Zeros: x = 0, x = -1

There are a total of 5 zeros to that equation, 2 of them are irrational.
The -1 and 0 are correct, work on the other 3. Try graphing to find the zeros (a graphing calculator or Excel will work nice for that).

The possible choices for the remaining rational factor would have integral factors of the first and last coefncients (the 2 and 15). After that you can apply the quadratic formula to find the remaining factors.

Yes, PolluxCastor is right.

But how to find the remaining Zeros?

Here is some help.

I note (unless my calculations are wrong)

y = 2x^3 - 5x^2 - 6x + 15 has values -9, 14, 6, -1, 6 respectively when x = -2, -1, 1, 2, 3

Therefore there have to be zeros between x = -2 and -1, between 1 and 2, and between 2 and 3.

Why? Because signs of values of y change between these values of x.

Using EXCEL for calculations, and using the above, I get approx. following Zeros:

1.732, -1.732, 2.5

Of course if you could have guessed that one factor was (x - 2.5) then other factors could be found by a quadratic equation.

Funny thing is, now that I know about 2.5, it is so very obvious that (x - 2.5) or (2x -5) is a factor!